Answer:
Explanation:
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Answer:
1.Hydrogenation of Alkenes and akynes.
2.Reaction of alkylhalides.
3. Halogenation.
Moles of methanol = 9.27x10^24/6.02x10^23 = 15.398 moles.
Mass of methanol = moles of methanol x molar mass of methanol
= 15.398 x 32.042
= 493.38 grams.
Hope this helps!
16 g. The mass of 0.60 mol Al is 16 g.
Molar mass of Al = 26.98 g/mol
Mass of Al = 0.60 mol Al x (26.98 g Al/1 mol Al) = 16 g Al
In this case, we are going to assume that there are 100 atoms to make things easier.
Let R% be the abundance of n-15. With this in mind, we calculate the abundance of n-14 to be 100%-R%
14.0031*(100-R)% + 15.001 * R%= 14.00674
In this case, we can delete or ignore the % sign since we do not want to carry it around, however, we need to keep in mind that the final answer is in %
14.0031*(100-R) + 15.001 * R= 14.00674
1400.31-14.0031R+15.001R=1400.674
0.9979R=0.364
R=0.3648
Then, the abundance of n-15 is 0.3648%
